All questions of Analog Electronics for Electrical Engineering (EE) Exam

To find the input resistance of the network, we can use the formula:

Input resistance (Rin) = (Voltage gain / Current gain) * (Feedback factor / Amplifier gain)

Given:
Amplifier gain (Av) = 100
Feedback factor (β) = 5

The voltage gain (Av) can be calculated as the ratio of output voltage (Vo) to input voltage (Vi):

Av = Vo / Vi

Since it is a voltage series feedback network, the output voltage (Vo) and input voltage (Vi) are related by the formula:

Vo = Vi / (1 + β * Av)

Substituting the given values:

Av = 100
β = 5

Vo = Vi / (1 + 5 * 100)
Vo = Vi / 501

Now, we can calculate the output voltage (Vo) using the formula:

Vo = Av * Vi

Vo = 100 * 4V
Vo = 400V

Substituting the calculated values into the equation for Vo:

400V = Vi / 501

Solving for Vi:

Vi = 400V * 501
Vi = 200,400V

Now, we can substitute the calculated values into the formula for input resistance (Rin):

Rin = (Av / Ai) * (β / Av)

Rin = (100 / (2mA / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = 1,000 * 0.05
Rin = 50

Therefore, the input resistance of the network is 50 ohms.

Both assertion and reason are individually correct statements. However, the reason for assertion is that due to the applied electric field, and electrostatic force is developed on the electron and the electrons would be accelerated in a direction opposite to the applied electric field and this motion is called directed motion of electron

81 .06%.

The parameters hfe and hte are commonly used to describe the characteristics of a transistor:

1. hfe: This is the current gain of the transistor, also known as the forward current transfer ratio (β). It represents the ratio of the collector current (IC) to the base current (IB). In this case, hfe = 100 means that for every 1 unit of base current, the collector current is 100 times larger.

2. hte: This is the reverse voltage transfer ratio, also known as the reverse current gain. It represents the ratio of the output current (IC) to the input voltage (VBE). In this case, hte = 5.2 k (kilo-ohms) means that for every 1 unit of input voltage, the output current is 5.2 k times larger.

These parameters are important in determining the operating characteristics and performance of a transistor in various circuits, such as amplifiers and switches.

An amplifier has an open loop voltage gain of "A." The open loop voltage gain, also known as the gain without any feedback applied, is a measure of how much the amplifier amplifies the input voltage. It is typically expressed as a ratio or in decibels.

The open loop voltage gain can be calculated by dividing the output voltage by the input voltage:

Open Loop Voltage Gain (A) = Output Voltage / Input Voltage

For example, if an amplifier has an output voltage of 10V and an input voltage of 1V, the open loop voltage gain would be 10.

It is important to note that the open loop voltage gain is not a fixed value and can vary depending on the frequency of the input signal. Amplifiers often have a frequency response curve that shows how the gain changes with frequency.

Additionally, the open loop voltage gain is usually much higher than the closed loop voltage gain when feedback is applied. Feedback is commonly used to stabilize the amplifier's performance and reduce distortion.

Theory.

Given data:
VGS = 0.5 V
IDSS = 10 mA
VDS = 5 V
gmo = 5 mA/V

To find:
Values of VP and ID

- Calculation of VP:
VP can be calculated using the formula:
VP = VGS - ID/2 * (1/gmo)

Given:
VGS = 0.5 V
IDSS = 10 mA
gmo = 5 mA/V

Substituting the values in the formula:
VP = 0.5 V - (10 mA)/(2 * 5 mA/V)
VP = 0.5 V - 1 V
VP = -0.5 V

- Calculation of ID:
ID can be calculated using the formula:
ID = IDSS * (1 - VGS/VP)^2

Given:
VGS = 0.5 V
IDSS = 10 mA
VP = -0.5 V

Substituting the values in the formula:
ID = 10 mA * (1 - 0.5 V / -0.5 V)^2
ID = 10 mA * (1 + 1)^2
ID = 10 mA * 4
ID = 40 mA

Therefore, the values of VP and ID for VGS = 0.5 V are:
VP = -0.5 V
ID = 40 mA

Hence, the correct answer is option C) 4 V and 7.65 mA.

Clipping Circuit and its Components
A clipping circuit is a circuit that controls the shape of a waveform by clipping or removing a portion of the applied wave. It is commonly used in signal processing and waveform generation applications.

A clipping circuit typically consists of three main components: an ordinary diode, a resistor, and a capacitor. These components work together to shape the waveform by selectively removing certain portions of the input signal.

Assertion (A): Clipping circuit controls the shape of the waveform by clipping or removing a portion of the applied wave.
This statement is true. A clipping circuit is designed to modify the shape of a waveform by removing or clipping certain portions of the input signal. By controlling the clipping threshold and the direction of the clipping (positive or negative), the circuit can shape the waveform according to the desired specifications.

Reason (R): It needs minimum three components namely an ordinary diode, a resistor and a capacitor.
This statement is false. While it is true that a clipping circuit commonly consists of an ordinary diode, a resistor, and a capacitor, it does not necessarily need all three components to function. The basic functionality of a clipping circuit can be achieved with just a diode and a resistor. The capacitor is optional and is used in certain applications to provide additional filtering or smoothing of the waveform.

Therefore, the correct answer is option 'C' - A is true, but R is false. The assertion (A) is true as clipping circuits do control the shape of the waveform by clipping or removing a portion of the applied wave. However, the reason (R) is false as a clipping circuit can function with just a diode and a resistor, and the presence of a capacitor is optional.

Given Data:
Nominal quiescent collector current = 1.2 mA
Range of β for the transistor = 80 to 120
Change in quiescent collector current = ±10%

Calculation:
1. Calculate the range of quiescent collector current:
Minimum quiescent collector current = 1.2 mA - 10% of 1.2 mA
= 1.2 mA - 0.12 mA
= 1.08 mA
Maximum quiescent collector current = 1.2 mA + 10% of 1.2 mA
= 1.2 mA + 0.12 mA
= 1.32 mA
2. Calculate the corresponding range for β:
Minimum β = 1.08 mA / 1.2 mA * 80
= 72
Maximum β = 1.32 mA / 1.2 mA * 120
= 132
3. Calculate the range in value for r:
Minimum value for r = 1 / 72
= 1.56 kΩ
Maximum value for r = 1 / 132
= 2.88 kΩ

Conclusion:
Therefore, the range in value for r is 1.56 kΩ to 2.88 kΩ. Hence, option D (1.56 kΩ to 2.88 kΩ) is the correct answer.

A FET tuned amplifier refers to an amplifier circuit that uses a field-effect transistor (FET) as the active device and is designed to amplify a specific frequency range.

Given the parameters:
- gm = 5 mA/V: This represents the transconductance of the FET, which is a measure of how much the drain current changes with respect to the gate-source voltage. In this case, the transconductance is 5 mA per volt.
- rd = 20 kΩ: This represents the drain-source resistance of the FET, which is also known as the output impedance. It is the resistance seen looking into the drain terminal with the source terminal shorted to ground.

To design a FET tuned amplifier, you would typically need additional information such as the desired frequency range, the gain requirements, input/output impedance requirements, etc. However, I can provide a general overview of the circuit.

A common configuration for a FET tuned amplifier is the common-source configuration, which provides high voltage gain and good input/output isolation. The key components in this configuration include:

1. FET: The FET is the active device that provides amplification. In this case, the FET has a transconductance of 5 mA/V and an output impedance of 20 kΩ.

2. Biasing Network: The FET requires proper biasing to operate in the active region. This typically involves using resistors and/or capacitors to provide the appropriate DC bias voltage at the gate and drain terminals.

3. Tuning Network: The tuning network consists of inductors and capacitors that are used to select the desired frequency range for amplification. These components create a resonant circuit that provides a narrow bandwidth around the desired frequency.

4. Coupling Capacitors: Coupling capacitors are used to block DC voltage and allow only the AC signal to pass between stages of the amplifier. They are typically placed at the input and output of the amplifier to couple the signals without interfering with the DC biasing.

5. Load Resistor: The load resistor is connected to the drain terminal of the FET and provides the output impedance of the amplifier. It is typically chosen to match the desired output impedance or to interface with the next stage of the circuit.

Overall, the specific design of a FET tuned amplifier will depend on the specific requirements and constraints of the application. It is important to consider factors such as gain, bandwidth, stability, and distortion when designing and implementing such an amplifier circuit.